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# For one toss of a certain coin, the probability that the out

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Manager
Joined: 16 Feb 2012
Posts: 200
Concentration: Finance, Economics
For one toss of a certain coin, the probability that the out [#permalink]

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09 Jul 2012, 07:50
14
00:00

Difficulty:

85% (hard)

Question Stats:

54% (01:36) correct 46% (01:37) wrong based on 322 sessions

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For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

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Math Expert
Joined: 02 Sep 2009
Posts: 46217
Re: For one toss of a certain coin, the probability that the out [#permalink]

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09 Jul 2012, 07:57
2
2
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

$$P(h)=0.6$$, so $$P(t)=0.4$$. We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: $$P(h=5)=0.6^5$$;

4h and 1t: $$P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4$$, multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, $$P(h\geq{4})=0.6^5+5*0.6^4*0.4$$.

Answer: E.
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Posts: 47
Concentration: General Management, Technology
GPA: 3.9
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Re: For one toss of a certain coin, the probability that the out [#permalink]

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19 Sep 2014, 20:00
Bunuel wrote:
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

$$P(h)=0.6$$, so $$P(t)=0.4$$. We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: $$P(h=5)=0.6^5$$;

4h and 1t: $$P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4$$, multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, $$P(h\geq{4})=0.6^5+5*0.6^4*0.4$$.

Answer: E.

Hi Bunuel,

Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is..
Please let me know.
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Math Expert
Joined: 02 Sep 2009
Posts: 46217
Re: For one toss of a certain coin, the probability that the out [#permalink]

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20 Sep 2014, 13:22
kpali wrote:
Bunuel wrote:
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is o.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

$$P(h)=0.6$$, so $$P(t)=0.4$$. We want to determine the probability of at least 4 heads in 5 tries.

At least 4 heads means 4 or 5. Let's calculate each one:

5 heads: $$P(h=5)=0.6^5$$;

4h and 1t: $$P(h=4)=\frac{5!}{4!}*0.6^4*0.4=5*0.6^4*0.4$$, multiplying by 5 as 4h and 1t may occur in 5 different ways:
hhhht
hhhth
hhthh
hthhh
thhhh

So, $$P(h\geq{4})=0.6^5+5*0.6^4*0.4$$.

Answer: E.

Hi Bunuel,

Thanks for the explanation. I am having difficulty understanding though when should we consider order to be important and when not? When it is not explicitly stated that is..
Please let me know.

Think logically, the case of 4 heads and 1 tail can occur in several ways (each of which having the same probability), so we have to account for the order here.
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Joined: 17 Dec 2012
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For one toss of a certain coin, the probability that the out [#permalink]

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31 May 2017, 23:17
1
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

1. Probability of getting 4 heads and one tail is 5(0.6)^4 * (0.4). 5 is nothing but 5C4 ways. of 4 heads happening.
2. Probability of getting heads all the 5 times is (0.6)^5
3. Total probability is answer E.
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Re: For one toss of a certain coin, the probability that the out [#permalink]

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25 Jun 2017, 12:29
1
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

Refer to the solution in the picture
Attachments

Solution Head.jpeg [ 48.14 KiB | Viewed 1197 times ]

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Posts: 70
Location: India
GMAT 1: 600 Q49 V21
WE: Engineering (Manufacturing)
Re: For one toss of a certain coin, the probability that the out [#permalink]

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08 Aug 2017, 07:46
mihir0710 wrote:
Stiv wrote:
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

A. (0.6)^5
B. 2(0.6)^4
C. 3(0.6)^4
D. 4(0.6)^4(0.4) + (0.6)^5
E. 5(0.6)^4(0.4) + (0.6)^5

Refer to the solution in the picture

Thanks for the lucid explanation..
In GMAT can we expect problems related to Bayes’ theorem?
Please suggest...
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Re: For one toss of a certain coin, the probability that the out   [#permalink] 08 Aug 2017, 07:46
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# For one toss of a certain coin, the probability that the out

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